In considering conditional equations, a natural question to ask is whether it is possible to solve an equation involving an unknown set, and if so, what types of solutions can be expected. We shall see that solution of equations is possible, but that in general the solution is not unique, but represents upper and lower boundsfor the unknown set.

Suppose that an equation involving an unknown set X is given and that all letters other than X which appear in the equation represent known sets. This equation may be written in the form p(X) = 0, where p(X)  represents a polynomial. If p(X) contains terms in which neither X nor X' appear, each such term may be multiplied by          X + X' to produce an equivalent equation of the form AX + BX' = 0. This equation is equivalent to the two equations AX = 0 and BX' = 0, or the condition that B ⊆ X ⊆ A'. Any set which satisfies this condition is a solution of the equation. B is termed the least solution and A' the greatest solution of the equation. Clearly, a solution exists if and only if B ⊆ A' or, equivalently, if and only if AB = 0. This equation is called the eliminant of X,  or the condition for consistency of the given equation.

Whether an equation is consistent or not will depend, in general, upon the meanings of the sets A and B. However, in some cases the condition AB = 0 may be satisfied (or cannot be satisfied) independently of the meanings of the sets. For example, the equation CDX + C'X' = 0 is consistent for arbitrary sets C and D, since (CD) (C') = 0 regardless of the meaning of C and D. Again, the equation (M + M')X + X' = 0 is inconsistent, regardless of the meaning of set M.

The solution of an equation is unique only in the case that B = A', in the notation used above. Primarily because solutions are rarely unique,  solving equations plays a very minor role in the algebra of sets. Artificial problems can be constructed, but they are of little practical importance.

EXAMPLE 1. Solve for X: BX = C.

Solution. This equation is equivalent to BC'X + B'C + CX' = 0, and if B'C is multiplied by X + X', we obtain (B'C + B(Y)X + CX' = 0. Hence the general solution is C ⊆X ⊆(B'C' + BC). The eliminant is (B'C + BC')C = 0,  or B'C = 0. That is, the equation is consistent provided C ⊆ B, and then any set between C and B'C' + BC satisfies the equation.

Not only is it possible to solve a single equation for an unknown set X,  but simultaneous equations or sets of conditions in any form can also be solved. Since we proved in Section (Conditional equations.) that any set of conditions is equivalent to a single equation, simultaneous equations always reduce to a single equation for which the method described above is adequate.

EXAMPLE 2. Determine the nature of set X if the following conditions are given: .AX'⊆B, B'X + C = 1.

Solution. The two given conditions reduce to the single equation AB'X' + BC' + C'X' = 0. This equation is equivalent to BC'X + (AB' + C')X' = 0.

Hence (A B' + C') ⊆ X ⊆ (B' + C) is the general solution, and the condition for consistency is BC' = 0. If the conditions are consistent, any set X containing AB' + C' will satisfy the conditions.

مواضيع ذات صلة

Zero Polynomial

Kronecker,s Polynomial Theorem

THE ALGEBRA OF SETS-The number of elements in a set

THE ALGEBRA OF SETS-Conditional equations

THE ALGEBRA OF SETS-Properties of set inclusion

THE ALGEBRA OF SETS- Expanding, factoring, and simplifying

THE ALGEBRA OF SETS-Fundamental laws

THE ALGEBRA OF SETS-Venn diagrams

THE ALGEBRA OF SETS-The combination of sets

THE ALGEBRA OF SETS-Element and set

THE ALGEBRA OF SETS-Introduction

SYMBOLIC LOGIC AND THE ALGEBRA OF PROPOSITIONS-Special problems